Respected All.
Suppose that $x,y$ are positive real numbers such that $4x+3y=1$. We have to find the greatest value of $(\frac{3x}{5})^2(\frac{5y}{4})^4$.
Now using weighted A.M. $\geqslant$ weighted G.M. on the positive real numbers $\frac{4x}{2}, \frac{3y}{4}$ with weights $2,4$ respectively we get \begin{align} \frac{2(\frac{4x}{2})+4(\frac{3y}{4})}{2+4}&\geqslant [(\frac{4x}{2})^2(\frac{3y}{4})^4]^{\frac{1}{2+4}}\\ \Rightarrow (\frac{1}{2})^6&\geqslant x^2y^4\frac{2^4 3^4}{2^{10}}\\ \Rightarrow x^2y^4&\leqslant \frac{1}{3^4}\\ \Rightarrow (\frac{3x}{2})^2(\frac{5y}{4})^4&\leqslant \frac{3^2 5^4}{2^2 4^4}\frac{1}{3^4}=\frac{5^4}{2^{10} 3^2} \end{align} Thus the greatest value of $(3x/2)^2(5y/4)^4$ is $\frac{5^4}{2^{10}3^2}$.
My question is, is the above approach is correct ? Or did I make any mistake ? please help me if there is any better way of doing this using weighted A.M. > weighted G.M.
Thank you in advance